What is the solution of this Analytical geometry question?

Question

Find and graph the equation of the straight line through the intersection point of the two lines: x+4y = 3 , x + y = 0 and parallel to the straight line passing through the points ( 2, -1 ) and ( 4, 6 ) 

Answer 1

Find the intersection point of the two given lines:

 

x + 4y = 3

x + y   = 0

----------------

 

Subtracting will eliminate the x variable

 

 x + 4y = 3

-(x + y = 0)

---------------

      3y = 3

        y = 1

 

Plugging into the line x+y= 0:

x + 1 = 0

x = -1

 

The lines cross at (-1, 1)

 

we want the equation of the line through (-1,1)

parallel to the line through (2,-1), (4,6)

 

Slope of that line is (6-(-1))/(4-2) = 7/2

Parallel lines have same slope so you want the

equation of line through (-1,1) with slope 7/2

 

y - y₁ = m(x -x₁) where (x₁, y₁) = (-1,1)

 

y - 1 = (7/2)(x-(-1))

y - 1 = (7/2)(x+1)

2(y-1) = 7(x+1)

2y-2 = 7x + 7

-9 = 7x - 2y

 

7x - 2y = -9

 

To graph plug in a value for x and solve for y

to find a point on the graph:

 

2y = 7x + 9

y = (7/2)x + 9/2

y = (7/2)x + 4 1/2

 

x     |   (7/2)x + 4 1/2

-----|----------------

0     |   4 1/2

-2   |   -2 1/2

 

There are two points on the line.

Plot and draw the line.

Answer 2

x + 4y = 3

x + y = 0

 

3y = 3

y = 1

x = -1

The intersection point is (-1,1)

 

The slope is (6 + 1)/(4 - 2) = 7/2

 

7/2 = (y - 1)/(x + 1)

7x + 5 = 2y is the required equation.